A) \[\frac{1}{15}\]
B) \[\frac{2}{15}\]
C) \[\frac{4}{15}\]
D) none of these
Correct Answer: A
Solution :
[a] \[{{\left( \frac{1}{x} \right)}^{r}}{{=}^{m}}{{C}_{r}}{{x}^{2m-3r}}=\frac{{{(15+1)}^{n}}}{15}\] \[=\frac{{{(}^{n}}{{C}_{0}}{{15}^{n}}{{+}^{n}}{{C}_{1}}{{15}^{n-1}}+...{{+}^{n}}{{C}_{n-1}}15{{+}^{n}}{{C}_{n}})}{15}\] = integer \[+\frac{1}{15}\] Hence, the fractional part of \[\frac{{{2}^{4n}}}{15}\]is \[\frac{1}{15}\]
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